QJEEM17Ch6d5e: JEE Main 2017 Chemistry - Class 11 JEEM17Ch6d5e Solution: 1 gram of a carbonate (M 2 CO 3 ) on treatment with excess HCl…

M₂CO₃ + 2HCl $$\stackrel{}{⟶}$$ 2MCl + CO₂ + H₂O

Here weight of M₂CO₃ = 1 gm

Let molar mass of M₂CO₃ = M

$$\therefore$$ No of moles of M₂CO₃ = $$\frac{1}{M}$$

For this reaction,

$$\frac{\frac{1}{M}}{1}=\frac{0.01186}{1}$$

$$\Rightarrow M$$ = $$\frac{1}{0.01186}$$ = 84.3 g mol^-1

NOTE :

For any reaction this will be always valid.

For this reaction, 2KClO₃ $$\stackrel{}{⟶}$$ 2KCl + 3O₂

We can write, $$\frac{n_{KCl{O}_3}}{2}=\frac{n_{KCl}}{2}=\frac{n_{O_2}}{3}$$

This means